Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I've come up with this question in trying to solve a vaguely related exercise:

If $x_n$ is any sequence of points in $\mathbb{R}^n$ with $x_n \longrightarrow 0$, is there a path $\gamma(t)$, $\gamma(0)=0$ that goes through all $x_n$ and which is differentiable at $t=0$ ?

Thank you.

share|cite|improve this question
up vote 4 down vote accepted

As worded, the answer is yes: take a space-filling path and attach a differentiable piece to it. But this is not what you meant to ask. The question could have been this: Is there a path $\gamma\colon (-1,1)\to\mathbb R^n$ such that (a) $v=\gamma'(0)$ exists and is nonzero; (b) $\gamma(t_n)=x_n$ where $t_n$ is a sequence that decreases to $0$?

Then the answer is no. Here is an example in two dimensions, with basis vectors $e_1,e_2$: $x_n=\frac{1}{n}e_1$ if $n$ is odd and $x_n=\frac{1}{n}e_2$ if $n$ is even. If $\gamma(t)/t\to v$ as $t\to 0$, then $\frac{\gamma(t)}{|\gamma(t)|}\to \frac{v}{|v|}$. However, $\frac{\gamma(t_n)}{|\gamma(t_n)|}$ alternates between $e_1$ and $e_2$, a contradiction.

share|cite|improve this answer

It's actually quite easy. All we need is a smooth function $\sigma : [0,1] \rightarrow [0,1]$ such that $\sigma|_{[0,1/3]} \equiv 0$ and $\sigma|_{[2/3,1]} \equiv 1$. There are numerous ways to construct this, by integrating and scaling a bump function for example.

Now that we have that, let $\gamma_n : [0,1] \rightarrow \mathbb{R}^n$ be a path that goes in a straight line from $x_n$ to $x_{n+1}$. Then the path defined by $\gamma(t) = \gamma_n(\sigma(2^{n+1} (t - (1 - 2^{-n})))$ for $t \in [1-2^{-n}, 1-2^{-n-1}]$ (assuming the first point is $x_0$) is smooth and does the trick. I don't even think you need the sequence to be convergent.

share|cite|improve this answer
A stickler would ask what $\gamma(1)$ is. This is why the sequence needs to converge. – mixedmath May 23 '12 at 6:14
Ah, indeed, you're right. Also, sorry, I realized just now my answer is very similar to yours (a bit shorter though). – Najib Idrissi May 23 '12 at 6:16
Is it obvious that this is differentiable at the end point? – Michael May 23 '12 at 19:32
@Michael: It's very probably not, but Weltschmerz only asked differentiability at $t=0$. – Najib Idrissi May 24 '12 at 7:37
As I understood the question time zero is the end time but I might be mistaken. Weltschmerz, care to make it more precise? – Michael May 24 '12 at 18:43

As written, this answer is a sort of intuitive yes. In fact, let's make the path almost everywhere smooth. All the necessary intuition is to believe that if we have three points with prescribed tangent vectors at each, then there is a smooth path between them (sort of - I suppose there's a bit more). There are a few ways to see this: you might restrict yourself to the plane containing those three points and perform a sort of quintic Hermite approximation (or closer, a Bezier curve-like approximation). It might be a bit better if you actually straighten out the path in a small neighborhood around each of the points (small enough to not intersect the other neighborhoods), which you can justify with a bump function style argument. This isn't necessary, but perhaps makes it easier to visualize, as then the assertion is that you can 'smoothly twist' a line to go through any point? (I'm a bit uncertain of your familiarity with these things).

Then the idea would be to start with a straight line segment through the origin, which is clearly 'differentiable' at the origin. Then smoothly connect the line segment to $x_1$. And then to $x_2$. And so on.

If you're a generalized path sympathizer, then you might let $\gamma$ traverse the path from $x_n$ to $x_{n+1}$ at time $t \in [n, n+1]$, with the idea that traversing paths back to back is still a path. Or perhaps you're a stickler, and then you'll have to do this in something like at time $t \in [1 - 2^{-n}, 1 - 2^{-(n+1)}]$, and traversing the $n$th path at $2^n$th rate (still a path, though). If in addition, we bound the $n$th path in some decreasing neighborhood of the origin (which should exist as $x_n \to 0$), I believe we should even be able to say that $\gamma(1) = 0$.

The idea here is that you can sort of piecemeal your way through so that the path is smooth on $t \in (0,1)$, with $\gamma (0) = 0$. And with this particular construction, $\lim_{t \to 0} \gamma'(t)$ is defined, and gives the direction of the original line segment. And the fact that $x_n \to 0$ should mean that our path should descend to $0$ as well, although there's not a chance in the world that it will be smooth at $t = 1$.

The resulting path will be a little bit wonky, perhaps, but this seems reasonable in principle.

share|cite|improve this answer
Thank you. I'm confused, though. Isn't smoothness at $t=1$ what we want? (See my comment on N.I's solution. I thank you for your patience). – Weltschmerz May 23 '12 at 17:26
@N.I Never mind my comment. I think I didn't ask the right question. I guess what I want is a path that takes decreasing values of $t$ into increasing (with respect to $n$) values of $x_n$. Your statement, in your answer, that there will be no smoothness at $t=1$ tells me it's impossible. What would be an example to prove it? I think I should have required my path to defined on an interval $(-\epsilon,\epsilon)$. Both your answer and N.I's are very helpful anyway, but feel free to edit my post or vote to close it. – Weltschmerz May 23 '12 at 17:35

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.